a,  7\(x\).(2\(x\) + 10) = 0

        \(\left[{}\begin{matrix}x=0\\2x+10=0\end{matrix}\right.\)

         \(\left[{}\begin{matrix}x=0\\2x=-10\end{matrix}\right.\)

         \(\left[{}\begin{matrix}x=0\\x=-10:2\end{matrix}\right.\)

         \(\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)

Vậy \(x\in\){-5; 0}

b, - 9\(x\) : (2\(x\) - 10) = 0

      - 9\(x\) = 0

           \(x\) = 0

c, (4 - \(x\)).(\(x\) + 3) = 0

    \(\left[{}\begin{matrix}4-x=0\\x+3=0\end{matrix}\right.\)

    \(\left[{}\begin{matrix}x=4\\x=-3\end{matrix}\right.\)

Vậy \(x\in\) {-3; 4}

d, (\(x\) + 2023).(\(x\) - 2024) = 0

    \(\left[{}\begin{matrix}x+2023=0\\x-2024=0\end{matrix}\right.\)

    \(\left[{}\begin{matrix}x=-2023\\x=2024\end{matrix}\right.\)

Vậy \(x\) \(\in\) {-2023; 2024}

=>|2x-10|=-2x+10

=>2x-10<=0

=>x<=5

a, 7\(x\).(2\(x\) + 10) =0

    \(\left[{}\begin{matrix}x=0\\2x+10=0\end{matrix}\right.\)

    \(\left[{}\begin{matrix}x=0\\2x=-10\end{matrix}\right.\)

     \(\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)

Vậy \(x\in\) {-5; 0}

b, -9\(x\) : (2\(x\) - 10) = 0

    9\(x\)                   = 0 

     \(x\)                    = 0 

c, (4 - \(x\)).(\(x\) + 3)  = 0

    \(\left[{}\begin{matrix}4-x=0\\x+3=0\end{matrix}\right.\)

    \(\left[{}\begin{matrix}x=4\\x=-3\end{matrix}\right.\)

Vậy \(x\in\) {-3; 4}

d, (\(x\) + 2023).(\(x\) - 2024) = 0

     \(\left[{}\begin{matrix}x+2023=0\\x-2024=0\end{matrix}\right.\)

     \(\left[{}\begin{matrix}x=-2023\\x=2024\end{matrix}\right.\)

Vậy \(x\in\) {-2023; 2024}

a) 2x+8≤ 0 

⇔2x≤-8

⇔x≤-4

b) 4x-7 ≥ 2x -5

⇔2x-12 ≥ 0

⇔2x≥12

⇔x≥6

c) (2x-8)(15-3x)>0

TH1: 2x-8>0 ⇒x>4

        15-3x>0⇒x<5 

TH2:  2x-8<0 ⇒x<4

        15-3x<0⇒x>5 (vô lí)

vậy 4<x<5

a)\(=>2x=-10=>x=-5\)

b)\(=>-2x=-5=>x=\dfrac{-5}{-2}=\dfrac{5}{2}\)

c)\(4-x=0=>x=4-0=4\)

d)\(=>2x=-1=>x=-\dfrac{1}{2}\)

e)\(=>x^2=-2\)=> x ko tồn tại

f)\(=>x\left(2+1\right)=0=>3x=0=>x=0\)

`#040911`

`a)`

`(2x - 1)^2 - (2x + 5)(2x + 1) = 10`

`\Leftrightarrow 4x^2 - 4x + 1 - (4x^2 + 12x + 5) = 10`

`\Leftrightarrow 4x^2 - 4x + 1 - 4x^2 - 12x - 5 = 10`

`\Leftrightarrow (4x^2 - 4x^2) - (4x + 12x) + (1 - 5) = 10`

`\Leftrightarrow -16x - 4 = 10`

`\Leftrightarrow -16x = 10 + 4`

`\Leftrightarrow -16x = 14`

`\Leftrightarrow x = \dfrac{-7}{8}`

Vậy, `x= \dfrac{-7}{8}`

`b)`

`9^2(x - 1) + 25(1 - x) = 0`

`\Leftrightarrow 9^2(x - 1) - 25(x - 1) = 0`

`\Leftrightarrow (x - 1)(9^2 - 25) = 0`

`\Leftrightarrow`\(\left[{}\begin{matrix}x-1=0\\9^2-5^2=0\end{matrix}\right.\)

`\Leftrightarrow`\(\left[{}\begin{matrix}x=1\\\left(9-5\right)\left(9+5\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\4\cdot14=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\56=0\left(\text{vô lý}\right)\end{matrix}\right.\\ \text{Vậy, x = 1}\)

`c)`

\(x^2+3x-4=0\)

`\Leftrightarrow x^2 + 4x - x - 4 = 0`

`\Leftrightarrow (x^2 - x) + (4x - 4) = 0`

`\Leftrightarrow x(x - 1) + 4(x - 1) = 0`

`\Leftrightarrow (x + 4)(x - 1) = 0`

\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x-1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-4\\x=1\end{matrix}\right.\\ \text{ Vậy, }x\in\left(-4;1\right)\)

Trong các phương trình sau, phương trình nào vô nghiệm:

A. x² - 2x + 2 = 0

Vì: x² - 2x + 2 = x² - 2.x.1 + 1 + 1 = (x - 1)² + 1 > 0

Chúc bạn học tốt@@

A. x² - 2x + 2 = 0

\(\Leftrightarrow\) (x - 1)² + 1 = 0

Vì (x - 1)² \(\ge\) 0 với mọi x nên (x - 1)² + 1 \(\ge\) 1 > 0 với mọi x

\(\Rightarrow\) Phương trình vô nghiệm

Vậy S = \(\varnothing\)

B. x² - 2x + 1 = 0

\(\Leftrightarrow\) (x - 1)² = 0

\(\Leftrightarrow\) x - 1 = 0

\(\Leftrightarrow\) x = 1

Vậy S = {1}

C. x² - 2x = 0

\(\Leftrightarrow\) x(x - 2) = 0

\(\Leftrightarrow\) x = 0 hoặc x - 2 = 0

\(\Leftrightarrow\) x = 0 và x = 2

Vậy S = {0; 2}

D. 2x - 10 = 2x - 10

\(\Leftrightarrow\) 2x - 10 - 2x + 10 = 0

\(\Leftrightarrow\) 0x = 0

\(\Rightarrow\) Phương trình có vô số nghiệm

Vậy phương trình vô nghiệm là A. x² - 2x + 2 = 0

Chúc bn học tốt!!

\(a,\left|2x+2\right|+10=2x\)

*TH1 : \(\left|2x+2\right|=2x+2\Leftrightarrow2x+2>0\Leftrightarrow x>-1\)

\(\Rightarrow2x+2+10=2x\)

\(\Leftrightarrow2x-2x=-10-2\)

\(\Leftrightarrow0x=-12\left(vô\cdot lý\right)\)

*TH2 :\(\left|2x+2\right|=-2x-2\Leftrightarrow-2x-2< 0\Leftrightarrow x>-1\)

\(\Rightarrow-2x-2+10=2x\)

\(\Leftrightarrow-2x-2x=-10+2\)

\(\Leftrightarrow-4x=-8\)

\(\Leftrightarrow x=\dfrac{1}{2}\left(nhận\right)\)

Vậy \(S=\left\{\dfrac{1}{2}\right\}\)

\(b,\left|x-6\right|=\left|3-2x\right|\)

\(\Leftrightarrow\left[{}\begin{matrix}x-6=3-2x\\x-6=-3+2x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)

Vậy \(S=\left\{-3;3\right\}\)